Function behavior of $e^x$ and $x^2$ so I'm doing a math problem requiring me to find the area between two curves, $e^x$ and $x^2$. It gave intervals ($x=-1$ and $x = 1$) but I wanted to find their intersection points so I can better envision/sketch the area I'm trying to obtain. However, I find that (at least on math computational websites) that I have to graph both functions to actually find where exactly they intersect. I graphed the function on geogebra to see what it looked like and it only has $1$ intersection point. For some reason, I was envisioning that it would have at least $2$ intersection points but in this case, $e^x$ travels faster than $x^2$. Now I'm thinking that my understanding of $e^x$'s behavior is not clear. Is there any way to deduce that the functions only cross once? I always find intersection points via just equating the functions together and solving for $x$, but it seems that in this case it cannot be done.
 A: The analytic solution for the crossover point is:
$$x = -2 W\left(\frac{1}{2}\right),$$
where $W$ is the Lambert's W function.
This confirms there is only a single real solution.

A: As David G. Stork answered, there is analytical solution for the zero of function
$$f(x)=e^x-x^2$$ but it involves special function (Lambert in this case).
By inspection or graphing, you noticed that the solution is between $-1$ and $0$. To find the zero, the simplest could be Newton method; starting for example at the midpoint, that is to say using $x_0=-\frac 12$, the calculations would converge very fast
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & -0.500000 \\
 1 & -0.721926 \\
 2 & -0.703601 \\
 3 & -0.703467
\end{array}
\right)$$
A: Here is how to find the intersection point, albeit approximately. Let $f(x) = e^2 -x^2$ with the root $r=-1+a$ and expand,
$$f(r)=0\approx f(-1)+f'(-1)a+\frac 12 f''(-1) a^2$$
Evaluate,
$$f(-1)=\frac 1e-1, \>\>\> f'(-1)=\frac 1e + 2,\>\>\>f''(-1)=\frac 1e -2$$
and we have,
$$0=1-e+(1+2e)a+\frac 12 (1-2e)a^2$$
Solve for the root,
$$r =\frac{\sqrt{10e-1}-2}{1-2e}=-0.703$$
